3.17 \(\int \frac{\sin ^3(x)}{(1-\cos (x))^2} \, dx\)

Optimal. Leaf size=12 \[ \cos (x)+2 \log (1-\cos (x)) \]

[Out]

Cos[x] + 2*Log[1 - Cos[x]]

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Rubi [A]  time = 0.0362866, antiderivative size = 12, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {2667, 43} \[ \cos (x)+2 \log (1-\cos (x)) \]

Antiderivative was successfully verified.

[In]

Int[Sin[x]^3/(1 - Cos[x])^2,x]

[Out]

Cos[x] + 2*Log[1 - Cos[x]]

Rule 2667

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{\sin ^3(x)}{(1-\cos (x))^2} \, dx &=\operatorname{Subst}\left (\int \frac{1-x}{1+x} \, dx,x,-\cos (x)\right )\\ &=\operatorname{Subst}\left (\int \left (-1+\frac{2}{1+x}\right ) \, dx,x,-\cos (x)\right )\\ &=\cos (x)+2 \log (1-\cos (x))\\ \end{align*}

Mathematica [A]  time = 0.0161087, size = 13, normalized size = 1.08 \[ \cos (x)+4 \log \left (\sin \left (\frac{x}{2}\right )\right )-1 \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[x]^3/(1 - Cos[x])^2,x]

[Out]

-1 + Cos[x] + 4*Log[Sin[x/2]]

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Maple [A]  time = 0.05, size = 11, normalized size = 0.9 \begin{align*} \cos \left ( x \right ) +2\,\ln \left ( -1+\cos \left ( x \right ) \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)^3/(1-cos(x))^2,x)

[Out]

cos(x)+2*ln(-1+cos(x))

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Maxima [A]  time = 2.09661, size = 14, normalized size = 1.17 \begin{align*} \cos \left (x\right ) + 2 \, \log \left (\cos \left (x\right ) - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)^3/(1-cos(x))^2,x, algorithm="maxima")

[Out]

cos(x) + 2*log(cos(x) - 1)

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Fricas [A]  time = 1.593, size = 47, normalized size = 3.92 \begin{align*} \cos \left (x\right ) + 2 \, \log \left (-\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)^3/(1-cos(x))^2,x, algorithm="fricas")

[Out]

cos(x) + 2*log(-1/2*cos(x) + 1/2)

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Sympy [B]  time = 0.645438, size = 61, normalized size = 5.08 \begin{align*} \frac{2 \log{\left (\cos{\left (x \right )} - 1 \right )} \cos{\left (x \right )}}{\cos{\left (x \right )} - 1} - \frac{2 \log{\left (\cos{\left (x \right )} - 1 \right )}}{\cos{\left (x \right )} - 1} + \frac{\sin ^{2}{\left (x \right )}}{\cos{\left (x \right )} - 1} + \frac{2 \cos ^{2}{\left (x \right )}}{\cos{\left (x \right )} - 1} - \frac{2 \cos{\left (x \right )}}{\cos{\left (x \right )} - 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)**3/(1-cos(x))**2,x)

[Out]

2*log(cos(x) - 1)*cos(x)/(cos(x) - 1) - 2*log(cos(x) - 1)/(cos(x) - 1) + sin(x)**2/(cos(x) - 1) + 2*cos(x)**2/
(cos(x) - 1) - 2*cos(x)/(cos(x) - 1)

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Giac [A]  time = 1.18609, size = 16, normalized size = 1.33 \begin{align*} \cos \left (x\right ) + 2 \, \log \left (-\cos \left (x\right ) + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)^3/(1-cos(x))^2,x, algorithm="giac")

[Out]

cos(x) + 2*log(-cos(x) + 1)